Mathematical Methods for Physics and Engineering - Matematica.NET

TENSORSThus, by inverting the matrix G in (26.60), we find that the elements g ij are given incylindrical polar coordinates byĜ =[g ij ]=⎛⎝ 1 0 0 1/ρ 00⎞⎠ . ◭0 0 1So far we have not considered the components of the metric tensor gj i with onesubscript and one superscript. By analogy with (26.56), these mixed componentsare given bygj i = e i · e j = δ j i ,and so the components of gji are identical to those of δj i.Wemaythereforeconsider the δj i to be the mixed components of the metric tensor g.26.16 General coordinate transformations and tensorsWe now discuss the concept of general transformations from one coordinatesystem, u 1 ,u 2 ,u 3 , to another, u ′1 ,u ′2 ,u ′3 . We can describe the coordinate transformusing the three equationsu ′i = u ′i (u 1 ,u 2 ,u 3 ),for i =1, 2, 3, in which the new coordinates u ′i can be arbitrary functions of the oldones u i rather than just represent linear orthogonal transformations (rotations)of the coordinate axes. We shall assume also that the transformation can beinverted, so that we can write the old coordinates in terms of the new ones asu i = u i (u ′1 ,u ′2 ,u ′3 ),As an example, we may consider the transformation from spherical polar toCartesian coordinates, given byx = r sin θ cos φ,y = r sin θ sin φ,z = r cos θ,which is clearly not a linear transformation.The two sets of basis vectors in the new coordinate system, u ′1 ,u ′2 ,u ′3 , are givenas in (26.55) bye ′ i = ∂r and e ′i = ∇u ′i . (26.66)∂u ′iConsidering the first set, we have from the chain rule that∂r∂u j = ∂u′ i∂r∂u j ∂u ′ i ,960